Fibonacci and -Lucas Numbers
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Hindawi Publishing Corporation Chinese Journal of Mathematics Volume 2013, Article ID 107145, 7 pages http://dx.doi.org/10.1155/2013/107145 Research Article Incomplete -Fibonacci and -Lucas Numbers José L. Ramírez Instituto de Matematicas´ y sus Aplicaciones, Universidad Sergio Arboleda, Colombia Correspondence should be addressed to JoseL.Ram´ ´ırez; [email protected] Received 10 July 2013; Accepted 11 September 2013 Academic Editors: B. Li and W. van der Hoek Copyright © 2013 JoseL.Ram´ ´ırez. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We define the incomplete k-Fibonacci and k-Lucas numbers; we study the recurrence relations and some properties of these numbers. 1. Introduction In [6],theauthorsdefinetheincompleteFibonacciandLucas -numbers. Also the authors define the incomplete bivariate Fibonacci numbers and their generalizations have many Fibonacci and Lucas -polynomials in [7]. interesting properties and applications to almost every field On the other hand, many kinds of generalizations of of science and art (e.g., see [1]). Fibonacci numbers are Fibonaccinumbershavebeenpresentedintheliterature.In defined by the recurrence relation particular, a generalization is the -Fibonacci Numbers. For any positive real number ,the-Fibonacci sequence, 0 =0, 1 =1, +1 = +−1,⩾1. (1) say {,}∈N, is defined recurrently by There exist a lot of properties about Fibonacci numbers. =0, =1, = + ,⩾1. In particular, there is a beautiful combinatorial identity to ,0 ,1 ,+1 , ,−1 (4) Fibonacci numbers [1] In [8], -Fibonacci numbers were found by studying ⌊(−1)/2⌋ −−1 = ∑ ( ). the recursive application of two geometrical transformations (2) =0 used in the four-triangle longest-edge (4TLE) partition. These numbers have been studied in several papers; see8 [ – From (2), Filipponi [2] introduced the incomplete Fibonacci 14]. () () numbers and the incomplete Lucas numbers .They For any positive real number ,the-Lucas sequence, say are defined by { } , ∈N, is defined recurrently by −1− −1 =2, =, = + . () = ∑ ( ) (=1,2,3,...;0≤≤⌊ ⌋) , ,0 ,1 ,+1 , ,−1 (5) 2 =0 If =1,wehavetheclassicalLucasnumbers.Moreover, = + ,⩾1 − , ,−1 ,+1 ;see[15]. () = ∑ ( ) (=1,2,3,...;0≤≤⌊ ⌋) . In [12], the explicit formula to -Fibonacci numbers is =0 − 2 ⌊(−1)/2⌋ −−1 (3) = ∑ ( )−2−1, , (6) =0 Further in [3], generating functions of the incomplete Fibonacci and Lucas numbers are determined. In [4], Djord- and the explicit formula of -Lucas numbers is jevic´ gave the incomplete generalized Fibonacci and Lucas ⌊/2⌋ − numbers. In [5], Djordjevic´ and Srivastava defined incom- () = ∑ ( )−2i. , (7) plete generalized Jacobsthal and Jacobsthal-Lucas numbers. =0 − 2 Chinese Journal of Mathematics From (6)and(7), we introduce the incomplete - Proof. Use Definition 1 to rewrite the right-hand side of (11) Fibonacci and -Lucas numbers and we obtain new recurrent as relations, new identities, and their generating functions. +1 − −−1 ∑ ( )−2 + ∑ ( )−2−1 =0 =0 2. The Incomplete -Fibonacci Numbers +1 − +1 − = ∑ ( )−2+1 + ∑ ( )−2+1 Definition 1. The incomplete -Fibonacci numbers are −1 defined by =0 =1 +1 − − (13) =−2+1 (∑ [( )+( )]) − +1 ( ) −1 −1 =0 −1− −2−1 −1 = ∑ ( ) , 0≤≤⌊ ⌋. (8) , 2 +1 =0 −+1 = ∑ ( )−2+1 −0 =0 In Table 1, some values of incomplete -Fibonacci num- =,+2. bers are provided. We note that Proposition 3. One has ⌊(−1)/2⌋ =. 1, (9) −−1 ∑ ( ) + =+ , 0≤≤ . ,+ ,+2 (14) =0 2 =1 For , we get incomplete Fibonacci numbers [2]. Proof (by induction on ). Sum (14)clearlyholdsfor=0and Some special cases of (8)are =1(see (11)). Now suppose that the result is true for all <+1;weproveitfor+1: 0 =−1; (≥1) , +1 +1 , ∑ ( )+ ,+ 1 −1 −3 =0 , = + (−2) ; (≥3) +1 + 2 −1 −3 (−4)(−3) −5 = ∑ [( )+( )] = + (−2) + ; (≥5) −1 ,+ , 2 =0 ⌊(−1)/2⌋ = ; (≥1) +1 +1 , , = ∑ ( )+ + ∑ ( )+ ,+ −1 ,+ =0 =0 ⌊(−3)/2⌋ { − ( ) = , even (≥3) . , { 2 −1 ( ) =+ +( )++1 +1 + ∑ ( )++1 +1 { , odd ,+2 +1 ,++1 ,++1 (10) =−1 =+ +0+∑ ( )++1 +1 +( ) ,+2 ,++1 −1 , =0 2.1. Some Recurrence Properties of the Numbers , Proposition 2. The recurrence relation of the incomplete - + ++1 =,+2 +∑ ( ),++1 +0 Fibonacci numbers , is =0 + ++1 =,+2 +,+2+1 +1 +1 −2 ,+2 =,+1 +,,0≤≤ . (11) ++1 2 =,+2+2. (15) The relation (11) canbetransformedintothenonhomoge- neous recurrence relation Proposition 4. For ≥2+2, −1 −1− −1−2 ∑ −1− =+1 −+1 . ,+2 =,+1 +, −( ) . (12) ,+ ,++1 ,+1 (16) =0 Chinese Journal of Mathematics 3 Table 1: The numbers ,,for1⩽⩽10. n\l 01 2 3 4 11 2 k 2 3 k2 +1 3 4 k3 +2 4 2 4 2 5 k4 +3 +3 +1 5 3 5 3 6 k5 +4 +4 +3 6 4 6 4 2 6 4 2 7 k6 +5 +5 +6 +5 +6 +1 7 5 7 5 3 7 5 3 8 k7 +6 +6 + 10 +6 + 10 +4 8 6 8 6 4 8 6 4 2 8 6 4 2 9 k8 +7 +7 + 15 +7 + 15 + 10 +7 + 15 + 10 +1 9 7 9 7 5 9 7 5 3 9 7 5 3 10 k9 +8 +8 + 21 +8 + 21 + 20 +8 + 21 + 20 +5 Proof(byinductionon). Sum (16) clearly holds for =1 By deriving into the previous equation (respect to ), it is (see (11)). Now suppose that the result is true for all <. obtained We prove it for : −1 − −−1 ∑,+ =∑,+ +,+ ⌊(−1)/2⌋ −1− =0 =0 + = ∑ (−2) ( )−2−1 , , =(+1 −+1 )+ =0 ,++1 ,+1 ,+ (17) (22) +1 +1 +1 =( + )− ⌊(−1)/2⌋ ,++1 ,+ ,+1 −1− = −2 ∑ ( )−2−1. , +1 +1 +1 =0 =,++2 − ,+1. Note that if ,in(4), is a real variable, then , =, and From Lemma 5, they correspond to the Fibonacci polynomials defined by 1 =0 { if () = =1 +1 { if (18) − () + () >1. +( , , ) { −1 if , 2 +4 Lemma 5. One has (23) ⌊(−1)/2⌋ +1 () − () +−1 () −1− () = = −2 ∑ ( )−2−1. 2 +4 , (19) =0 () − () = . 2 +4 See Proposition 13 of [12]. From where, after some algebra (20)isobtained. Lemma 6. One has Proposition 7. One has ⌊(−1)/2⌋ 2 −1− (( +4)−4), −, ∑ ( )−1−2 = . 2 =0 2( +4) (20) 4 + { , , { ( even) Proof. From (6)wehavethat ⌊(−1)/2⌋ { 2(2 +4) { ∑ , = { 2 (24) ⌊(−1)/2⌋ −1− {( +8) + = ∑ ( ) −2. =0 { , , ( ) . , (21) 2 odd =0 { 2( +4) 4 Chinese Journal of Mathematics Table 2: The numbers ,,for1⩽⩽9. n\l 01 2 3 4 1 k 2 2 2 +2 3 3 3 +3 4 4 2 4 2 4 +4 +4 +2 5 5 3 5 3 5 +5 +5 +5 6 6 4 6 4 2 6 4 2 6 +6 +6 +9 +6 +9 +2 7 7 5 7 5 3 7 5 3 7 +7 +7 + 14 +7 + 14 +7 8 8 6 8 6 4 8 6 4 2 8 6 4 2 8 +8 +8 + 20 +8 + 20 + 16 +8 + 20 + 16 +2 9 9 7 9 7 5 9 7 5 3 9 7 5 3 9 +9 +9 + 27 +9 + 27 + 30 +9 + 27 + 30 +9 ⌊(−1)/2⌋ Proof −1 −1− = ∑ (⌊ ⌋+1)( )−1−2 ⌊(−1)/2⌋ =0 2 ∑ , =0 ⌊(−1)/2⌋ −1− − ∑ ( )−1−2 0 1 ⌊(−1)/2⌋ =0 =, +, +⋅⋅⋅+, ⌊(−1)/2⌋ −1−0 −1 −1− −1−2 =( )−1 =(⌊ ⌋+1) − ∑ ( ) . 0 , 2 =0 −1−0 −1−1 (25) +[( )−1 +( )−3] 0 1 From Lemma 6,(24)isobtained. [ [ −1−0 −1−1 3. The Incomplete -Lucas Numbers +⋅⋅⋅+[ ( )−1 +( )−3 [ 0 1 Definition 8. The incomplete -Lucas numbers are defined by [ − = ∑ ( )−2, 0≤≤⌊ ⌋. −1 , − 2 (26) −1−⌊ ⌋ =0 2 ] −1−2⌊(−1)/2⌋] +⋅⋅⋅+( −1 ) ] In Table 2,somenumbersofincomplete-Lucas numbers ⌊ ⌋ ] 2 are provided. ] We note that −1 −1−0 ⌊/2⌋ = . =(⌊ ⌋+1)( )−1 1, (27) 2 0 Some special cases of (26)are −1 −1−1 +⌊ ⌋( )−3 0 =; (≥1) , 2 1 , 1 = +−2; (≥2) , −1 , −1−⌊ ⌋ 2 (−3) −1−2⌊(−1)/2⌋ 2 = +−2 + −4; (≥4) , +⋅⋅⋅+( −1 ) , 2 ⌊ ⌋ (28) 2 ⌊/2⌋ , =,; (≥1) , ⌊(−1)/2⌋ −1 −1− , −2 ( even) = ∑ (⌊ ⌋+1−)( )−1−2 ⌊(−2)/2⌋ ={ (≥2) . , =0 2 , − ( odd) Chinese Journal of Mathematics 5 3.1. Some Recurrence Properties of the Numbers , Proposition 12. One has Proposition 9. One has − ∑ ( )+ =+ ,0≤≤ . ,+ ,+2 2 (37) =−1 + ,0≤≤⌊⌋. =0 , ,−1 ,+1 2 (29) Proof. Using (29)and(14), we write Proof. By (8), rewrite the right-hand side of (29)as −1 −2− − ∑ ( )+ = ∑ ( )[+−1 ++ ] ∑ ( )−2−2 + ∑ ( )−2 ,+ ,+−1 ,++1 =0 =0 =0 =0 −1− − = ∑ ( )+−1 + ∑ ( )+ (38) = ∑ ( )−2 + ∑ ( )−2 ,+−1 ,++1 −1 =0 =0 =1 =0 =−1+ ++ =+ . −1− − −1 (30) ,−1+2 ,+1+2 ,+2 = ∑ [( )+( )] −2 −( ) −1 −1 =0 − Proposition 13. For ≥2+1, = ∑ ( )−2 +0 − =0 −1 −1− +1 +1 ∑,+ =,++1 −,+1. (39) =,. =0 The proof can be done by using (31) and induction on . Proposition 10. The recurrence relation of the incomplete - Lemma 14. One has Lucas numbers , is ⌊/2⌋ − −2 +1 +1 ∑ ( ) = [, −,]. (40) ,+2 =,+1 +,, 0≤≤⌊ ⌋. (31) − 2 2 =0 The relation (31) canbetransformedintothenonhomoge- The proof is similar to Lemma 6. neous recurrence relation − Proposition 15. One has = + − ( )−2. ,+2 ,+1 , (32) − ⌊/2⌋ { {, + , ( even) Proof. Using (29)and(11), we write ∑ = 2 , {1 (41) =0 ( + ) ( ) . +1 +1 {2 , , odd ,+2 =,+1 +,+3 −1 +1 Proof. An argument analogous to that of the proof of =, +,−1 +,+2 +,+1 (33) Proposition 7 yields +1 −1 =(, +,+2)+,−1 +,+1 ⌊/2⌋ ⌊/2⌋ − ∑ =(⌊ ⌋+1) − ∑ ( )−2. =+1 + . , 2 , − (42) ,+1 , =0 =0 From Lemma 14,(41)isobtained. Proposition 11. One has −1 4. Generating Functions of the Incomplete = −−2 , 0≤≤⌊ ⌋. , ,+2 ,−2 2 (34) -Fibonacci and -Lucas Number Proof. By (29), In this section, we give the generating functions of incomplete -Fibonacci and -Lucas numbers. = −−1,−2 =−1 −−1, ,+2 ,+1 , ,−2 ,−1 , (35) ∞ Lemma 16 (see [3,page592]).Let {}=0 be a complex whence, from (31), sequence satisfying the following nonhomogeneous recurrence −2 −1 relation: ,+2 −,−2 =,+1 −,−1 =,. (36) =−1 +−2 + (>1) , (43) 6 Chinese Journal of Mathematics where and are complex numbers and {} is a given complex Now let sequence.